/*! This file is auto-generated */ .wp-block-button__link{color:#fff;background-color:#32373c;border-radius:9999px;box-shadow:none;text-decoration:none;padding:calc(.667em + 2px) calc(1.333em + 2px);font-size:1.125em}.wp-block-file__button{background:#32373c;color:#fff;text-decoration:none} Problem 3 A flat bar of width \(b\) and th... [FREE SOLUTION] | 91Ó°ÊÓ

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A flat bar of width \(b\) and thickness \(t\) has a hole of diameter \(d\) drilled through it (see figure). The hole may have any diameter that will fit within the bar. What is the maximum permissible tensile load \(P_{\max }\) if the allowable tensile stress in the material is \(\sigma_{t} ?\)

Short Answer

Expert verified
The maximum tensile load is given by \( P_{\max} = \sigma_t \times (b \times t - \frac{\pi}{4}d^2) \).

Step by step solution

01

Define the Available Area

The available area for stress calculation, after the hole is drilled, is the original area minus the area of the hole. The original cross-sectional area of the bar is given by \( A_0 = b \times t \). The area of the hole is the area of a circle with diameter \( d \), which is \( A_h = \frac{\pi}{4}d^2 \). The net cross-sectional area is thus \( A_{net} = A_0 - A_h = b \times t - \frac{\pi}{4}d^2 \).
02

Apply the Tensile Stress Formula

The maximum permissible tensile load \( P_{\max} \) can be found using the permissible tensile stress \( \sigma_t \). The tensile stress is defined as \( \sigma_t = \frac{P_{\max}}{A_{net}} \). Thus, we can rearrange this formula to find \( P_{\max} \): \[ P_{\max} = \sigma_t \times A_{net} = \sigma_t \times (b \times t - \frac{\pi}{4}d^2) \].
03

Substitute Known Values (if any)

If the values for \( b \), \( t \), \( d \), and \( \sigma_t \) are known, substitute them into the equation to find the numerical value of \( P_{\max} \). Otherwise, this formula expresses the general relationship between these variables.

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Cross-Sectional Area
The cross-sectional area in a material is crucial for understanding how it will respond to different forces. Imagine cutting a slice through the object and looking at the face of the cut; the cross-sectional area is the size of this face. For a flat bar with a width (\( b \)) and thickness (\( t \)), the original cross-section is really just a rectangle. You calculate it by multiplying width and thickness: \( A_0 = b \times t \).
But if you add a hole, we lose some of that area! A hole reduces the cross-section because it takes away material. For example, a circle hole of diameter (\( d \)) burrows through the bar and removes area that equals \( A_h = \frac{\pi}{4}d^2 \).
To find the net cross-sectional area (\( A_{net} \)) that is left for handling stress after drilling, subtract the hole's area from the original area: \( A_{net} = b \times t - \frac{\pi}{4}d^2 \). This net area is what's available to bear loads, making it essential for calculating tensile stress.
Tensile Stress
Tensile stress occurs when a material, like our flat bar, is pulling apart. Imagine pulling on the ends of a piece of taffy; that's tensile stress at work! It measures how much force is acting over an area. Mathematically, tensile stress (\( \sigma_t \)) is represented as the force (\( P \)) applied divided by the cross-sectional area (\( A_{net} \)) over which it acts: \( \sigma_t = \frac{P}{A_{net}} \).
Think of stress as the effort an area is enduring under a load. A smaller cross-section means greater stress because the same force acts over less area. Conversely, a larger cross-section lowers stress because that force is spread out more broadly. When calculating with a drilled bar, use the net area (\( A_{net} \)), not the original, as material is missing. Tensile stress helps designers and engineers ensure that materials do not fail under loads and are safe to use.
Allowable Stress
Allowable stress is the maximum stress a material can endure without failure. In engineering, safety is paramount, so knowing this limit prevents disastrous scenarios. The "allowable" part implies that engineering standards have set a safe threshold within which the material must operate.
You find allowable stress values (\( \sigma_{t} \)) in materials charts, and it's used to ensure designs are safe. By multiplying the allowable stress with the net cross-sectional area (\( A_{net} \)), we determine the maximum permissible load (\( P_{\max} \)) the material can handle:\( P_{\max} = \sigma_t \times A_{net} \).
This equation ensures your material won't experience stress beyond what it's designed for. Always remember, when designing something to bear a load, consider not just the theoretical maximum but what is prudently "allowable" to ensure safety and longevity of your structure or component.

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Most popular questions from this chapter

A circular steel rod \(A B\) of diameter \(d=15 \mathrm{mm}\) is stretched tightly between two supports so that initially the tensile stress in the rod is \(60 \mathrm{MPa}\) (see figure). An axial force \(P\) is then applied to the rod at an intermediate location \(C\) (a) Determine the plastic load \(P_{P}\) if the material is elastoplastic with yield stress \(\sigma_{Y}=290 \mathrm{MPa}\) (b) How is \(P_{P}\) changed if the initial tensile stress is doubled to 120 MPa?

An aluminum pipe has a length of \(60 \mathrm{m}\) at a temperature of \(10^{\circ} \mathrm{C} .\) An adjacent steel pipe at the same temperature is \(5 \mathrm{mm}\) longer than the aluminum pipe. At what temperature (degrees Celsius) will the aluminum pipe be \(15 \mathrm{mm}\) longer than the steel pipe? (Assume that the coefficients of thermal expansion of aluminum and steel are \(\alpha_{a}=23 \times 10^{-6 / 0} \mathrm{C}\) and \(\alpha_{s}=12 \times 10^{-6 /^{\circ} \mathrm{C}}\) respectively.)

Two identical bars \(A B\) and \(B C\) support a vertical load \(P\) (see figure). The bars are made of steel having a stress-strain curve that may be idealized as elastoplastic with yield stress \(\sigma_{Y}\) Each bar has cross-sectional area \(A\) Determine the yield load \(P_{Y}\) and the plastic load \(P_{P^{-}}\)

Two cables, each having a length \(L\) of approximately \(40 \mathrm{m},\) support a loaded container of weight \(W\) (see figure). The cables, which have effective cross-sectional area \(A=48.0 \mathrm{mm}^{2}\) and effective modulus of elasticity \(E=160 \mathrm{GPa},\) are identical except that one cable is longer than the other when they are hanging scparately and unloaded. The difference in lengths is \(d=100 \mathrm{mm}\). The cables are made of steel having an elastoplastic stressstrain diagram with \(\sigma_{Y}=500\) MPa. Assume that the weight \(W\) is initially zero and is slowly increased by the addition of material to the container. (a) Determine the weight \(W_{Y}\) that first produces yielding of the shorter cable. Also, determine the corresponding elongation \(\delta_{Y}\) of the shorter cable. (b) Determine the weight \(W_{P}\) that produces yielding of both cables. Also, determine the elongation \(\delta_{P}\) of the shorter cable when the weight \(W\) just reaches the value \(W_{P}\) (c) Construct a load-displacement diagram showing the weight \(W\) as ordinate and the elongation \(\delta\) of the shorter cable as abscissa. (Hint: The load displacement diagram is not a single straight line in the region \(0 \leq W \leq W_{Y^{\prime}}\)

A steel cable with nominal diameter \(25 \mathrm{mm}\) (see Table \(2-1\) ) is used in a construction yard to lift a bridge section weighing \(38 \mathrm{kN},\) as shown in the figure. The cable has an effective modulus of elasticity \(E=140 \mathrm{GPa}\) (a) If the cable is \(14 \mathrm{m}\) long, how much will it stretch when the load is picked up? (b) If the cable is rated for a maximum load of \(70 \mathrm{kN}\) what is the factor of safety with respect to failure of the cable?

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